Find the maximum height before the ladder slips: slipping ladder problem general solution .

In this static equilibrium ladder problem a person climbs a ladder, and we compute the maximum height before the ladder slips. 🧠 Access full flipped physics courses with video lectures and examples at https://www.zakslabphysics.com/ We find the slipping ladder problem general solution, given the mass of the ladder, the mass of the person, the angle of incline of the ladder and the coefficient of static friction at the bottom of the ladder as purely symbolic quantities. We begin with a force diagram for the ladder. We start with the force of gravity acting downward on the center of mass of the ladder. Next, we include the force of gravity downward at the location of the person on the ladder (the force downward on the ladder is equal to the weight of the person). The ladder rests against a smooth wall which is only capable of producing a horizontal normal force, and the bottom end of the ladder has a friction force and normal force. We argue carefully by analyzing forces in the x direction that the static friction force at the bottom of the ladder must point to the right. We take a moment to analyze qualitatively why the ladder slips if the person climbs too high: the person's weight vector is moving farther up the ladder which increases the lever arm for the clockwise torque with respect to a rotation axis at the bottom of the ladder. This requires an increase in the normal force exerted by the smooth wall, which in turn requires an increase in the static friction force at the bottom of the ladder. Eventually, the person can climb high enough to exceed the maximum static friction force, and the ladder will slip. Next, we need to compute the perpendicular component of each force acting on the ladder. We have to use our prior knowledge that the angle between the perpendicular and the vertical is the same as the angle of incline to get the perpendicular component of weight for the ladder and the person. We have to use alternate interior angles to find the perpendicular component of the normal force exerted by the wall. Now we're ready to apply the equations of static equilibrium. We start with the equation for the net torque. The net torque with respect to the chosen rotation axis (the contact point with the ground) must vanish, and that means the clockwise torque equals the counterclockwise torque. Given the perpendicular components of each force acting on the ladder, we quickly write down an equation expressing the fact that the sum of the torques is zero. Next, we write down an equation for balancing forces in the horizontal direction, and finally, we write an equation balancing forces in the vertical direction. We now have a system of three equations and three unknowns, where the unknowns are the two normal forces exerted by the wall and the floor and the unknown distance x : the location of the person at the moment the ladder slips. We solve this system by using substitution and isolate x to find the maximum height of the person before the ladder slips. We arrange our final solution with the length of the ladder factored out, and a term in parentheses in which every part is unitless. This allows us to verify that the units of the final answer are indeed length units, and we're done!